[php] Get the year from specified date php

I have a date in this format 2068-06-15. I want to get the year from the date, using php functions. Could someone please suggest how this could be done.

You can try strtotime() and date() functions for output in minimum code and using standard way.

echo date('Y', strtotime('2068-06-15'));

Assuming you have the date as a string (sorry it was unclear from your question if that is the case) could split the string on the - characters like so:

$date = "2068-06-15";
$split_date = split("-", $date);
$year = $split_date[0];

You can achieve your goal by using php date() & explode() functions:

$date = date("2068-06-15");

$date_arr = explode("-", $date);

$yr = $date_arr[0];

echo $yr;

That is it. Happy coding :)

I would use this:

$parts = explode('-', '2068-06-15');
echo $parts[0];

It appears the date is coming from a source where it is always the same, much quicker this way using explode.

You can use the strtotime and date functions like this:

echo date('Y', strtotime('2068-06-15'));

Note however that PHP can handle year upto 2038

You can test it out here

If your date is always in that format, you can also get the year like this:

$parts = explode('-', '2068-06-15');
echo $parts[0];

You wrote that format can change from YYYY-mm-dd to dd-mm-YYYY you can try to find year there

$parts = explode("-","2068-06-15");
for ($i = 0; $i < count($parts); $i++)
{
     if(strlen($parts[$i]) == 4)
     {
          $year = $parts[$i];
          break;
      }
  }

$Y_date = split("-","2068-06-15");
$year = $Y_date[0];

You can use explode also

  public function getYear($pdate) {
    $date = DateTime::createFromFormat("Y-m-d", $pdate);
    return $date->format("Y");
}

public function getMonth($pdate) {
    $date = DateTime::createFromFormat("Y-m-d", $pdate);
    return $date->format("m");
}

public function getDay($pdate) {
    $date = DateTime::createFromFormat("Y-m-d", $pdate);
    return $date->format("d");
}

<?php
list($year) = explode("-", "2068-06-15");
echo $year;
?>